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17 tháng 1

a) \(\left(\dfrac{3}{29}-\dfrac{1}{5}\right)\cdot\dfrac{29}{3}\)

\(=\dfrac{3}{29}\cdot\dfrac{29}{3}-\dfrac{1}{5}\cdot\dfrac{29}{3}\)

\(=1-\dfrac{29}{15}\)

\(=\dfrac{15-29}{15}\)

\(=-\dfrac{14}{15}\) 

b) \(\dfrac{3}{4}\cdot\dfrac{7}{9}+\dfrac{1}{4}\cdot\dfrac{7}{9}\)

\(=\dfrac{7}{9}\cdot\left(\dfrac{1}{4}+\dfrac{3}{4}\right)\)

\(=\dfrac{7}{9}\cdot1\)

\(=\dfrac{7}{9}\)

c) \(\dfrac{1}{7}\cdot\dfrac{5}{9}+\dfrac{5}{9}\cdot\dfrac{1}{7}+\dfrac{5}{9}\cdot\dfrac{3}{7}\) 

\(=\dfrac{5}{9}\cdot\left(\dfrac{1}{7}+\dfrac{1}{7}+\dfrac{3}{7}\right)\)

\(=\dfrac{5}{9}\cdot\dfrac{5}{7}\)

\(=\dfrac{25}{63}\)

d) \(4\cdot11\cdot\dfrac{3}{4}\cdot\dfrac{9}{121}\)

\(=\left(4\cdot\dfrac{3}{4}\right)\cdot\left(11\cdot\dfrac{9}{121}\right)\)

\(=3\cdot\dfrac{9}{11}\)

\(=\dfrac{27}{11}\)

12 tháng 10 2019

A=\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{49.51}\)

=\(\frac{2}{1}-\frac{2}{3}+\frac{2}{3}-\frac{2}{5}+\frac{2}{5}-\frac{2}{7}+...+\frac{2}{49}-\frac{2}{51}\)

\(2.(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51})\)

=2.\((1-\frac{1}{51})\)

=\(2.\frac{50}{51}\)

=\(\frac{100}{51}\)

25 tháng 8 2018

Bài này mình làm xong rồi nhưng lỡ tay bấm nút hủy.

MONG CÁC BẠN  

25 tháng 8 2018

\(\frac{-1}{2}-\frac{-3}{5}+\frac{-1}{9}+\frac{1}{131}-\frac{-2}{7}+\frac{4}{35}-\frac{7}{18}\)

\(=\frac{-1}{2}+\frac{3}{5}-\frac{1}{9}+\frac{1}{131}+\frac{2}{7}+\frac{4}{35}-\frac{7}{18}\)

\(=\frac{1}{10}-\frac{122}{1179}+\frac{2}{5}-\frac{7}{18}\)

\(=-\frac{41}{11790}+\frac{1}{90}\)

\(=\frac{1}{131}\)

học tốt

24 tháng 6 2018

......................?

mik ko biết

mong bn thông cảm 

nha ................

−12−−35+−19+1131−−27+435−718

=−12+35−19+1131+27+435−718

=110−1221179+25−718

=−4111790+190

Bài 1: 

\(=\dfrac{-1}{2}+\dfrac{3}{5}-\dfrac{1}{9}+\dfrac{1}{131}+\dfrac{2}{7}+\dfrac{4}{35}-\dfrac{7}{18}\)

\(=\left(-\dfrac{1}{2}-\dfrac{1}{9}-\dfrac{7}{18}\right)+\left(\dfrac{3}{5}+\dfrac{2}{7}+\dfrac{4}{35}\right)+\dfrac{1}{131}\)

\(=\dfrac{-9-2-7}{18}+\dfrac{21+10+4}{35}+\dfrac{1}{131}\)

=1/131

Bài 2:

b: \(B=\dfrac{1}{99}-\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{98\cdot99}\right)\)

\(=\dfrac{1}{99}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{98}-\dfrac{1}{99}\right)\)

\(=\dfrac{1}{99}-\dfrac{98}{99}=-\dfrac{97}{99}\)

Ta có: \(A=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)\cdot...\cdot\left(1-\dfrac{1}{9}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot\dfrac{4}{5}\cdot...\cdot\dfrac{8}{9}\)

\(=\dfrac{1}{9}\)